What are the first three non-zero terms of the Maclaurin series for the function $f(x)=\sin(2x)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ 2x-\frac{2x^3}{3!}+\frac{2x^5}{5!}$ (Choice B) B $ x-\frac{8x^3}{3!}+\frac{32x^5}{5!}$ (Choice C) C $ x+\frac{8x^3}{3!}-\frac{32x^5}{5!}$ (Choice D) D $ 2x-\frac{8x^3}{3!}+\frac{32x^5}{5!}$
Start with the Maclaurin series for $\sin (x)$. $\sin (x)=x-\frac{{{x}^{3}}}{3!}+\frac{{{x}^{5}}}{5!}-...+{{\left( -1 \right)}^{n}}\frac{{{x}^{2n+1}}}{\left( 2n+1 \right)!}+...$ Substitute $2x$ for $x$. $\begin{aligned} &\phantom{=}\sin (2x) \\\\ &=2x-\dfrac{(2x)^3}{3!}+\dfrac{(2x)^5}{5!}-...+{{\left( -1 \right)}^{n}}\dfrac{{{(2x)}^{2n+1}}}{\left( 2n+1 \right)!}+... \\\\ &=2x-\dfrac{8x^3}{3!}+\dfrac{32x^5}{5!}-...+{{\left( -1 \right)}^{n}}\dfrac{{{(2x)}^{2n+1}}}{\left( 2n+1 \right)!}+... \end{aligned}$ The first three non-zero terms are $2x-\dfrac{8x^3}{3!}+\dfrac{32x^5}{5!}$.